Htykuut

I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = frac{(a_n)^2}{10n + 3} + 1$

I know that intuitively the limit will be $1$, since $frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.

Hints would be appreciated.

First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.

It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$

Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.

It should be clear that $a_n >0$ for all n.

Now prove, by induction, that $(a_n)$ is decreasing.

Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.